Timeline for Why is it so hard to give examples of differentially closed fields?
Current License: CC BY-SA 4.0
11 events
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| Mar 4, 2022 at 20:41 | comment | added | user44143 | David Marker's introduction to the model theory of differential fields, especially the first and last two pages, may be helpful here for anyone unfamiliar with this subject: library.msri.org/books/Book39/files/dcf.pdf | |
| Mar 4, 2022 at 0:20 | comment | added | Erik Walsberg | I don't think there is anything like that in the transseries book, but maybe it's possible to do some kind of formal power series construction. | |
| Mar 3, 2022 at 18:21 | comment | added | Gro-Tsen | At least the existence of the algebraic closure of $\mathbb{C}(t)$ follows from ZF, since it is the relative algebraic closure of $\mathbb{C}(t)$ in the field of Puiseux series (which is algebraically closed by a theorem of Newton), so no axiom of choice is needed here. This is why it would be interesting to know if the existence of its differential closure follows from ZF. | |
| Mar 3, 2022 at 15:40 | comment | added | Zerox | @Wojowu Thanks for the correction! Do you have any result about the constructibility of differential closed field in $ZF$? According to your statement arbitrary existence is not provable, but how about the particular case of $\Bbb{C}(t)$? | |
| Mar 3, 2022 at 15:29 | comment | added | Wojowu | @Zerox That's not correct. Existence of algebraic closures is known to not be equivalent to AC. It is also known that existence of algebraic closures in general is not provable in ZF alone, though it is provable for a lot of fields one encounters in practice, like subfields of $\mathbb C$ or fields of rational functions over them. | |
| Mar 3, 2022 at 15:05 | comment | added | Zerox | I'm afraid there is no valid answer, because even in the algebraically closed case the equivalence between $AC$ and the existence of such closed field is still open, see this question. Special case do exist like $\Bbb{C} \cong \Bbb{R}[i]$ can be proven to be algebraically closed without $AC$, however topological properties are required. | |
| Mar 3, 2022 at 14:55 | comment | added | Gro-Tsen | @Zerox Are you claiming, for example, that it is consistent with ZF (no Choice) that $\mathbb{C}(t)$ has no differential closure? Because that (if correct) would certainly answer my question! | |
| Mar 3, 2022 at 14:07 | comment | added | Zerox | Similar complexity caused by '$t$' occurs when completing the value-at-infinity order of real rational functions. | |
| Mar 3, 2022 at 14:06 | comment | added | Zerox | The complexity can be seen from that, in order to construct a nontrivial differential operator, we must introduce some transcendental element $t$ out of nowhere to the usual number field. | |
| Mar 3, 2022 at 14:01 | comment | added | Zerox | The constuction in the paper is equipping the algebraic closure of some infinite rational function space $\Bbb{Q}(\Lambda_t)$ with a "directional derivative". I believe the index set $T$ can be chosen better to avoid using orderings in the construction, but it is scarcely possible to prove the closedness under differential equations without using choice. | |
| Mar 3, 2022 at 13:09 | history | asked | Gro-Tsen | CC BY-SA 4.0 |